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Suppose A $ \in \mathbb C^{m\times m}$ has an $SVD: A = U\sum V^*$. Find an eigenvalue decomposition form of the $2m \times 2m$ hermitian matrix $$ B=\begin{bmatrix} 0&A^* \\ A&0 \end{bmatrix} $$

I cannot get the eigenvalue decomposition form of $ B=X\sum X^*$. How to do that?

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We have $$ B=\begin{pmatrix}0&V\\U&0\end{pmatrix}\begin{pmatrix}0&\Sigma\\\Sigma^*&0\end{pmatrix}\begin{pmatrix}0&U^*\\V^*&0\end{pmatrix}. $$ and $$ W\begin{pmatrix}0&\Sigma\\\Sigma^*&0\end{pmatrix}W^*=2\begin{pmatrix}(\Sigma\Sigma^*)^{1/2}&0\\0&-(\Sigma^*\Sigma)^{1/2}\end{pmatrix}, $$ where $$ W=\begin{pmatrix}(\Sigma^*)^{1/2}&(\Sigma^*)^{-1/2}\\ \Sigma^{1/2}&-\Sigma^{-1/2}\end{pmatrix}. $$ It remains to compute $W^{-1}$.

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  • $\begingroup$ In this form we can get $B=\begin{bmatrix} a&0 \\ 0&b \end{bmatrix}$ But in the problem the B is in the form $B=\begin{bmatrix} 0&a \\ b&0 \end{bmatrix}$ $\endgroup$ – Gatsby Sep 10 '16 at 20:18
  • $\begingroup$ I have corrected the answer. Sorry for possible misprints, not enough time to check. $\endgroup$ – Vladimir Sep 10 '16 at 20:53
  • $\begingroup$ How do you get the second row? Is this a trick or just by trial and error? $\endgroup$ – Gatsby Sep 10 '16 at 22:45
  • $\begingroup$ The eigenvalues of $\begin{pmatrix}0&\Sigma\\\Sigma&0\end{pmatrix}$ are those of $\begin{pmatrix}(\Sigma^*\Sigma)^{1/2}&0\\0&-(\Sigma^*\Sigma)^{1/2}\end{pmatrix}$ (this is easy to check by inspecting the characteristic polynomial, because $\Sigma$ is diagonal). In the one-dimensional case, it is also easy to find $W$ reducing our matrrix to this form. The general case follows readily. $\endgroup$ – Vladimir Sep 11 '16 at 4:52

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